Find the percentage increase in the area of a triangle if each s | Math Question and Answer

Answer:

$100(n^2 -1) \%$

Step by Step Explanation:

Consider a $\triangle QRS$ with sides $a, b$ and $c.$ Let $S = \dfrac { a+b+c } {2}$ .
Area of $\triangle QRS, A_1 = \sqrt { S(S-a)(S-b)(S-c) }$
Increasing the side of each side by $n$ times, we get a new $\triangle XYZ$ .
$\triangle XYZ$ has sides $na, nb$ and $nc$ .
By Heron's formula:
Area of new triangle $= \sqrt{ S_{1}(S_{1}-na)(S_{1}-nb)(S_{1}-nc) }$
Where, $S_1 = \dfrac { na + nb + nc }{2} = n × \dfrac { a+b+c } {2}$
$\begin{align} \text { Area of } \triangle XYZ & = \sqrt{ nS(nS-na)(nS-nb)(nS-nc) } \\ & = \sqrt{ n^{4}S(S-a)(S-b)(S-c)) } \\ & = n^2 \times A_1 \end{align}$
Increase in area $= n^2 A_1 - A_1$
$\%$ Increase in area $= \dfrac{ A_1(n^2 - 1) }{ \dfrac{ A_1 }{ 100 } } =$ $100(n^2 -1) \%$ .

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